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Question Number 168144 by cortano1 last updated on 04/Apr/22

$$\underset{x\to 0}{\lim }\frac{x^3\sqrt{1+x}-\sin {}^{3}x-\frac{1}{2}{x}^3\tan x}{(\sqrt[3]{1+2x^3}-1)\ln (1+x^2)}=?$$

Answered by qaz last updated on 05/Apr/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^3\sqrt{1+\mathrm{x}}-\sin {}^3\mathrm{x}-\frac{1}{2}{\mathrm{x}}^3\tan \mathrm{x}}{(\sqrt[3]{1+{2\mathrm{x}}^3}-1)\ln (1+{\mathrm{x}}^2)}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^3(1+\frac{1}{2}\mathrm{x}-\frac{1}{8}{\mathrm{x}}^2+...)-{\mathrm{x}}^3{(1-\frac{1}{6}{\mathrm{x}}^2...)}^3-\frac{1}{2}{\mathrm{x}}^3(\mathrm{x}+...)}{\frac{2}{3}{\mathrm{x}}^5}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(1+\frac{1}{2}\mathrm{x}-\frac{1}{8}{\mathrm{x}}^2+...)-(1+\left(\begin{array}{c}3\\ 1\end{array}\right)(-\frac{1}{6}{\mathrm{x}}^2)+...)-\frac{1}{2}(\mathrm{x}+...)}{\frac{2}{3}{\mathrm{x}}^2}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(-\frac{1}{8}+\frac{1}{2}){\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^2\right)}{\frac{2}{3}{\mathrm{x}}^2}$$$$=\frac{9}{16}$$

Commented by cortano1 last updated on 05/Apr/22

$$ans=\frac{9}{16}$$

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